Your question is related to Pell equations by the following result:
Theorem. Fix $m \in \mathbb{N}$. The values of $k$ for which $f(m, km)$ is a perfect square are exactly those of the form
$$ k = \frac{i-1}{2m} $$
where $i$ comes from any solution $(i, j)$ to the instance of Pell's equation given by
$$ i^2 - 4m(m+1)j^2 = 1. \tag{1} $$
That is, given $m$, the valid $k$ correspond exactly to solutions of an instance of Pell's equation with parameter dependent on $m$. Intuitively one might expect this to be the case by extrapolating from $m = 1$, where (as noted by Geoffrey Trang) solutions correspond to square-triangular numbers which in turn correspond to solutions to a specific Pell's equation.
After proving this Theorem, we'll use the theory linking Pell equations and Chebyshev polynomials to prove your conjecture about the $\sinh$ sequence.
We'll need the following technical lemma, which I originally posted as a conjecture on MSE in the process of solving your problem.
Lemma. Suppose $k(m+1)(km+1)$ is a perfect square. Then $(m+1) \mid k(km+1)$.
Proof. If $m+1$ is squarefree, then this follows from the fact that primes in perfect squares occur in even multiplicities. The general case is highly nontrivial; see the beautiful proof by Mastrem using Vieta jumping on my MSE post. $\square$
Proof of the Theorem
The core argument below is due to a very clever friend of mine who shared her solution with me offline; I'm just writing it up.
Proof. Fix $m \in \mathbb{N}$. We first prove that, if $k$ is such that $f(m, km)$ is a perfect square, then it must be the case that
$$ k = \frac{i-1}{2m} $$
where $i$ is a solution to $i^2 - 4m(m+1)j^2 = 1$. Conversely, we then show that for any such $i$, taking $k$ as above yields a positive integer such that $f(m, km)$ is a perfect square.
$(\Longrightarrow)$ Let $k \in \mathbb{N}$ be arbitrary such that
$$ f(m, km) = \frac{m^2k(m+1)(km+1)}{4} $$
is a perfect square. Then $k(m+1)(km+1)$ must also be a perfect square, so write
$$ k(m+1)(km+1) = s^2 \quad\text{for some $s \in \mathbb{N}$}. \tag{2} $$
Expanding (2) and rewriting as a quadratic in $k$, we obtain
$$ k^2m^2 + km + k^2m + k = s^2 \implies m(m+1)k^2 + (m+1)k - s^2 = 0. $$
The quadratic equation gives
$$ k = \frac{-(m+1) + \sqrt{(m+1)^2 + 4m(m+1)s^2}}{2m(m+1)} $$
where we can ignore the negative branch of the $\pm$ since $k$ is a positive integer by assumption. Now, consider the square root of the discriminant $\sqrt{D} = \sqrt{(m+1)^2 + 4m(m+1)s^2}$. Since $k$ is a positive integer, we see that $\sqrt{D}$ must be a positive integer divisible by $(m+1)$. Thus take $i \in \mathbb{N}$ such that
$$ \sqrt{D} = \sqrt{(m+1)^2 + 4m(m+1)s^2} = i(m+1), \tag{3} $$
and note that we can express
$$ k = \frac{-(m+1) + \sqrt{D}}{2m(m+1)} = \frac{-(m+1) + i(m+1)}{2m(m+1)} = \frac{i-1}{2m}. $$
Now, squaring both sides of Equation 3 yields
$$ (m+1)^2 + 4m(m+1)s^2 = i^2(m+1)^2. $$
This almost looks like a Pell's equation! To get it in the right form, apply the Lemma to $k(m+1)(km+1) = s^2$ to get $m+1 \mid k(km+1)$, so that $(m+1)^2 \mid s^2$ and hence $m+1 \mid s$. Then put $s = j(m+1)$ for some $j \in \mathbb{N}$; we get
$$ (m+1)^2 + 4m(m+1)j^2(m+1)^2 = i^2(m+1)^2. $$
Dividing through by $(m+1)^2$ on both sides and rearranging gives the true Pell's equation
$$ 1 + 4m(m+1)j^2 = i^2 \implies i^2 - 4m(m+1)j^2 = 1. $$
This argument shows that, for any $k$ such that $f(m, km)$ is a perfect square, there exist $(i, j)$ such that $k = (i-1)/(2m)$ and $i^2 - 4m(m+1)j^2 = 1$.
$(\Longleftarrow)$ Next, we prove that, given any such $i$, taking $k = (i-1)/(2m)$ yields a positive integer such that $f(m, km)$ is a perfect square. Let $(i, j)$ be a solution to
$$ i^2 - 4m(m+1)j^2 = 1. $$
We need to show that $k = (i-1)/(2m)$ is an integer, and that $m^2k(m+1)(km+1)/4$ is a perfect square. Ignore $m^2/4$ for the moment and substitute $(i-1)/(2m)$ for $k$ in $k(m+1)(km+1)$; direct computation shows that
$$ k(m+1)(km+1) = \frac{i-1}{2m}(m+1)\left(\frac{i-1}{2m}m + 1\right) = (i^2 - 1)\frac{m+1}{4m}. $$
On the other hand, multiplying both sides of $i^2 - 4m(m+1)j^2 = 1$ by $(m+1)$ yields
\begin{align*}
&i^2(m+1) - 4m(m+1)^2j^2 = m+1 \\
&\quad\implies (j(m+1))^2 = (i^2 - 1)\frac{m+1}{4m}
\end{align*}
where, per the previous computation, the right side is exactly $k(m+1)(km+1)$. Hence
$$ k(m+1)(km+1) = (j(m+1))^2 $$
is a perfect square, and moreover
$$ f(m, km) = \frac{m^2k(m+1)(km+1)}{4} = \left(\frac{m(m+1)j}{2}\right)^2 $$
is an integer since one of $m$, $m+1$ is divisible by $2$, and certainly is a perfect square.
It remains only to justify that $k = (i-1)/(2m)$ is always a positive integer. Recall that $i$ is a solution to $i^2 - 4m(m+1)j^2 = 1$, which is a Pell's equation with fundamental solution $(i_1, j_1) = (2m+1, 1)$. The theory of Pell equations tells us that the sequence of solutions $(i_\ell, j_\ell)$ is generated by the recurrence
$$ i_\ell = (2m+1)i_{\ell - 1} + 4m(m+1)j_{\ell -1}. $$
Working mod $2m$, the second term vanishes and we are left with $i_\ell \equiv i_{\ell - 1} \pmod{2m}$. A trivial induction argument then shows that every solution $i$ satisfies
$$ i \equiv i_1 \equiv 2m+1 \equiv 1 \pmod{2m} $$
implying that $2m \mid (i-1)$. We therefore conclude that $k = (i-1)/(2m)$ is always an integer, which finishes the proof. $\square$
Connections to Chebyshev polynomials and A082649
In the OP, a relationship to https://oeis.org/A082649, the triangle of coefficients in the expansion of $\sinh^2(nx)$ in powers of $\sinh(x)$, is conjectured. In this section, I discuss how everything is connected.
Chebyshev polynomials. We first link the sequence of $k$ to Chebyshev polynomials. For fixed $m$, let $k_n$ denote the $n$th smallest value of $k$ for which $f(m, km)$ is a perfect square. It follows from the Theorem that $k_n = (i_n - 1)/(2m)$, where $i_n$ is the $n$th solution of the Pell's equation $i^2 - 4m(m+1)j^2 = 1$. On the other hand, it is known that the Chebyshev polynomials $T_n$, $U_n$ manifest as solutions of a Pell's equation of the same form; in particular, they satisfy $$ T_n(x)^2 - (x^2 - 1)U_{n-1}(x)^2 = 1. $$
Setting $x = 2m+1$ so that $4m(m+1) = x^2 - 1$, it is straightforward to show that $i_n = T_n(2m+1)$.
Substituting into $k_n = (i_n - 1)/(2m)$ we thus see that, for fixed $m$, the $n$th smallest value of $k$ for which $f(m, km)$ is a perfect square is given by
$$ k_n = \frac{T_n(2m+1) - 1}{2m} $$
where $T_n$ is the $n$th Chebyshev polynomial of the first kind.
The $S_r$ sequence. Let me remark that from this expression for $k_n$, we easily obtain a connection to another interesting sequence on OEIS. In particular, https://oeis.org/A092184 defines a family of sequences $S_r$ given by
$$ S_r(n) = 2\frac{T_n((r-2)/2) - 1}{r-4}. $$
Setting $r = 4m+4$, we obtain
$$ S_{2m+4}(n) = 2\frac{T_n(2m+1) - 1}{4m} = \frac{T_n(2m+1) - 1}{2m} $$
which is precisely $k_n$! That is, for fixed $m$, the sequence of $k$ that work is exactly the member of the family of sequences $S_r$ with $r = 4m+4$.
The sinh sequence. Finally, let us prove the connection to https://oeis.org/A082649. The observation in the OP is that, given $m$, the $n$th smallest value of $k$ appears related to the coefficients of the expansion of $\sinh^2(nx)$ in powers of $\sinh(x)$. For instance, given $n = 3$, we have the expansion
$$ \sinh^2(3x) = 16\sinh^6 x + 24\sinh^4 x + 9\sinh^2 x. $$
If we replace each occurrence of $\sinh^2 x$ with $m$ and shift the powers down, we obtain $16m^2 + 24m + 9$ which happens to correspond to the $n=3$ smallest value of $k$.
It is not hard to see that a rigorous formulation of this claim is that
$$ k_n = \frac{T_n(2m+1) - 1}{2m} \stackrel{?}{=} \frac{\sinh^2(n\sinh^{-1}(\sqrt{m}))}{m}. $$
To prove this identity, set $t := \sinh^{-1}(\sqrt{m})$ so that $m = \sinh^2 t$. Then, working from the left side, we see that
\begin{align*}
\frac{T_n(2m+1) - 1}{2m} &= \frac{T_n(2\sinh^2 t + 1) - 1}{2m} \\
&= \frac{T_n(\cosh(2t)) - 1}{2m} \\
&= \frac{\cosh(2nt) - 1}{2m} \\
&= \frac{2\sinh^2(nt)}{2m} \\
&= \frac{\sinh^2(n\sinh^{-1}\sqrt{m})}{m}
\end{align*}
precisely as claimed, where we used standard hyperbolic trig identities on lines 2 and 4, and the Chebyshev polynomial identity $T_n(\cosh(\theta)) = \cosh(n\theta)$ on line 3. This concludes the proof of the connection to https://oeis.org/A082649, and resolves the original (very cute) problem in full.
